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To summarize the relation in one line, the Hadamard Transform is essentially the Quantum Fourier Transform for the special case of a single qubit. Traditionally, the Quantum Fourier Transform $U_{FT}$ for $n$ qubits is defined as $$U_{FT}|x\rangle_n = \frac{1}{2^{n/2}} \sum_{y=0}^{2^n - 1} e^{2 \pi i x y / 2^n}|y\rangle_n$$ However, this is not really the ...


3

We can think of the set $\mathbb{Z}_{2^n}$ as the set of all binary sequences of length $n$ and we can write it as $$ \mathbb{Z}_{2^n} = S^n_0 \cup S^n_1\tag1 $$ where $S^n_b=\{(y',b) |y\in\mathbb{Z}_{2^{n-1}}\}$ is the set of binary sequences of length $n$ that end in $b$, for $b\in\{0,1\}$. Note that $S_0\cap S_1=\emptyset$. Moreover, if $y\in S^n_0$ then $...


1

The classical Fourier transform acts on a vector $(x_{0}, x_{1}, \ldots, x_{N-1}) \in {C} ^{N}$ and maps it to the vector $(y_{0}, y_{1}, \ldots, y_{N-1}) \in {C} ^{N}$. According to the formula: $y_{k}={\frac {1}{\sqrt {N}}}\sum _{n=0}^{N-1}x_{n}\omega _{N}^{-kn},\quad k=0,1,2,\ldots ,N-1$ Similarly, the quantum Fourier transform acts on a quantum state $|x\...


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